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A set of generators for a topolgy that is not a basis?

  1. Sep 26, 2007 #1

    quasar987

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    It is clear that a basis for a topology generates that topology. (i.e. the smallest topology containing the basis is the topology that the basis is a basis of)

    But if we look at a generating set for a topology, is this generating set necessarily a basis?

    I am guessing that no in general (although it is true that the set of "generalized open rectangles" is both a generating set and a basis for the product topology) because otherwise a topology would always be determined by simply the union of its generators.

    But I have little experience with general topology and since this is just for personnal curiosity, I don't want to waste more time trying to find an example where the generating set is not a basis.

    Anybody's got an example?
     
  2. jcsd
  3. Sep 26, 2007 #2

    morphism

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    What is your definition of a generating set for a topology?
     
  4. Sep 27, 2007 #3

    quasar987

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    Let X be a set and G be a subset of 2^X (the power set). Then the topology of X generated by G is the intersection of all topologies containing G. It is therefor the smallest topology containing G.
     
  5. Sep 27, 2007 #4

    matt grime

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    You appear to have just answered your own question. Or at least your two definitions imply that a generating set is a basis.
     
  6. Sep 27, 2007 #5

    quasar987

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    I have not defined 'basis'. I did not think it was necessary, but I noticed that there are many different definitions around.. the one I use is this elegant one: If T is a topology for X, then a basis for T is a subset T' of T such that every open can be written as a reunion of elements of T'.
     
  7. Sep 28, 2007 #6

    matt grime

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    That is equivalent to a generating set, too by your own definition. I think you're on the wrong track of thinking of basis and spanning set from linear algbra.
     
  8. Sep 28, 2007 #7

    quasar987

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    Can you show me why?

    The objection I have is that another way to look at the topology generated by a set is that it is the topology you get by taking all the unions of the sets in your generating set, then taking all the intersections, and weird mix of intersections and unions.

    But if every generating set was a basis, we would not have to take the intersections and the weird mix, since by def of a basis, the only unions would suffice to create a topology.
     
  9. Sep 28, 2007 #8

    morphism

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    But basis elements are themselves open sets, so the intersection of finitely many of them is again an open set, and hence is a union of some collection of basis elements.
     
  10. Sep 29, 2007 #9

    quasar987

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    I guess this is the piece I was missing !

    Cheers morphism!
     
  11. Sep 29, 2007 #10

    Hurkyl

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    Let X = {1, 2, 3}.
    Consider the set G = { {1, 2} , {2, 3} }

    The smallest topology containing G is
    { {}, {2}, {1, 2}, {2, 3}, {1, 2, 3} }
    {2} isn't a union of elements of G, so G is certainly not a basis for the topology it generates.
     
  12. Sep 29, 2007 #11

    quasar987

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    Yeah! Hurkyl to the rescue! :cool:
     
  13. Sep 29, 2007 #12
    It is the subbase of the topology that generates the topology in the sense as you were talking about generating it.
     
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